I am trying to understand the proof of $\mathrm{E}[xy]=n(n-1)p_1p_2$ where x,y have a trinomial distribution with pmf:
$p(x,y) = \frac{n!}{x!y!(n−x−y)!}p_1^xp_2^yp_3^{n−x−y}$
The proof has the following steps
(1) $\mathrm{E}[xy]=\sum_{x=0}^n\sum_{y=0}^{n-x}xy\frac{n!}{x!y!(n−x−y)!}p_1^xp_2^yp_3^{n−x−y}$
(2) $\mathrm{E}[xy]=n(n-1)p_1p_2\sum_{x-1=0}^{n-1}\sum_{y-1=0}^{n-1-(x-1)}\frac{n-2!}{(x-1)!(y-1)!((n−2)-(x-1)−(y-1))!}p_1^{x-1}p_2^{y-1}p_3^{(n−2)-(x-1)−(y-1)}$
I understand that the point is to transform the double sum around $(x-1)$, $(y-1)$ meaning we pull out the $n(n-1)p_1p_2$ so that sum in (2) should equal 1 (by the multinomial theorem) but would have thought for that to happen the sum should be:
$\sum_{x-1=0}^{n-2}\sum_{y-1=0}^{n-2-(x-1)}\frac{n-2!}{(x-1)!(y-1)!((n−2)-(x-1)−(y-1))!}p_1^{x-1}p_2^{y-1}p_3^{(n−2)-(x-1)−(y-1)}$
I guess this is a trivial point to be hung up on but I can't seem to get past it.
Consider the terms where $x=n-1$, then the inner sum is $\sum_{y=0}^0 xy(something)=0$ because $xy=0$. Similarly, all the terms with $y=n-1$ are $0$, so the last index doesn't matter.